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mars1129 [50]
2 years ago
14

Pleaseee helppp with thisss asappp

Mathematics
1 answer:
Hoochie [10]2 years ago
8 0

We are given with a equation of Circle and we need to find it's radius and it's equation in standard form . But , let's recall , the standard equation of a circle is {\bf (x-h)^{2}+(y-k)^{2}=r^{2}} where <em>(h,k)</em> is the centre of the circle and radius is <em>r</em> . Proceeding further ;

{:\implies \quad \sf x^{2}+12y+22x+y^{2}-167=0}

Collecting <em>x</em> terms , y terms and transposing the constant to RHS ;

{:\implies \quad \sf (x^{2}+22x)+(y^{2}+12y)=167}

Now , as in standard equation their is a whole square , so we need to develop a whole square in LHS , for which we will use completing the square method , as coefficient of x² and y² is 1 , so adding 121 and 36 to LHS and RHS .

{:\implies \quad \sf (x^{2}+22x+121)+(y^{2}+12y+36)=167+121+36}

{:\implies \quad \sf (x+11)^{2}+(y+6)^{2}=324\quad \qquad \{\because a^{2}+2ab+b^{2}=(a+b)^{2}\}}

{:\implies \quad \bf \therefore \quad \underline{\underline{\{x-(-11)\}^{2}+\{y-(-6)\}^{2}=(18)^{2}}}}

On comparing this with the standard equation , we got our centre at <em>(-11,-6)</em> and radius is <em>18 units </em>

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